Question

If the radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$, then

• A. $4gf+8g-3f-6=0$
• B. $4gf-8g+3f+6=0$
• C. $4gf-8g-3f-6=0$
• D. $4gf-8g-3f+6=0$

Answer 1

The correct option is D $4gf-8g-3f+6=0$

The radical axis of the given circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+\left(\frac{3}{2}\right)x+4y+c=0$, is
$(2g-\frac{3}{2})x+(2f-4)y=0\Rightarrow (4g-3)x+4(f-2)y=0\cdots \left(1\right)$

This radical axis $\left(1\right)$ touches the circles $x^2+y^2+2x+2y+1=0\cdots \left(2\right)$

Distance from centre $(-1,-1)$ on the radical axis is equal to radius
$\left|\frac{(4g-3)(-1)+4(f-2)(-1)}{\sqrt{\left[\right(4g-3{)}^2+16(f-2{)}^2]}}\right|=\sqrt{(1+1-1)}\Rightarrow \left[\right(4g-3)+4(f-2){]}^2=(4g-3{)}^2+16(f-2{)}^2\Rightarrow 8(4g-3\left)\right(f-2)=0\therefore 4gf-8g-3f+6=0$